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# Steve gets on the elevator at the 11th floor of a building a

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Steve gets on the elevator at the 11th floor of a building a [#permalink]

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01 Apr 2006, 21:43
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Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

A. 19
B. 28
C. 30
D. 32
E. 44
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Mar 2014, 05:02, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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01 Apr 2006, 23:54

Steve travels at a rate of 57 floors/60s~1floor/1s

Joyce travels at a rate of 63 floors/60s~1floor/1s

Between them there are approximatelly 40 floors, so they will both have to travel ~ 20 floors to cross each other.... 20 floors is ~ 20 s

We need to use this formula.... Rate*Time=Distance

Steve: 57/60*20=19

Joyce: 63/60*20=21....

Adding things up we get 30......
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02 Apr 2006, 00:12
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Yeah C

Usuing Speed Formula: Speed = Distance / Time

Distance to be covered = 51-11 = 40

Speed of approach = 57 + 63 floors/min

Time = 40/120 = 1/3

So Steve will cover 57x 1/3 floors in that time = 19

So he will be in 19 + 11 floor = 30th Floor
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02 Apr 2006, 18:32
yeap its 30 i kinda figured it out ina different way little later because it was bothering me...

but its pretty much the same
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Re: Rate Problem... Really good question [#permalink]

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22 Mar 2014, 12:58
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]

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23 Mar 2014, 19:58
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Distance between Steve & Joyce = 40 floors

Lets say they meet on x floor

Time required for Joyce to reach x floor = Time required for Steve

So the equation would be

$$\frac{x}{57} = \frac{40-x}{63}$$

x = 19

Start is at 11th floor, so 11 + 19 = 30 = Answer = C
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]

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08 Aug 2015, 04:04
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]

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08 Sep 2016, 13:53
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]

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08 Sep 2016, 22:18
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

40 floors/120 floors per minute=1/3 minutes
11+57/3=30
51-63/3=30
C. 30
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Re: Steve gets on the elevator at the 11th floor of a building a [#permalink]

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08 Sep 2016, 23:05
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AshikaP wrote:
Steve gets on the elevator at the 11th floor of a building and rides up at a rate of 57 floors per minute. At the same time Joyce gets on an elevator on the 51st floor of the same building and rides down at a rate of 63 floors per minute. If they continue traveling at these rates, at which floor will their paths cross?

A. 19
B. 28
C. 30
D. 32
E. 44

Even though this question uses elevators (which usually means a tougher question), it involves very simple relative speed calculation.

Steve is on 11th floor and Joyce on 51st floor so they have a distance of 40 floors between them.
Steve's speed is 57 floors/min and Joyce's speed is 63 floors/min in opposite direction (towards Steve).
Their relative speed will be the sum of their individual speeds = 57 + 63 = 120 floors/min

Time taken to meet = 40/120 = 1/3 min = 20 secs

Distance covered by Steve in 20 secs = Speed*Time = 57 * (1/3) = 19 floors.
So he is at 11+19 = 30th floor when they meet.

Check this post on relative speed:
http://www.veritasprep.com/blog/2012/07 ... elatively/
http://www.veritasprep.com/blog/2012/08 ... -concepts/
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Re: Steve gets on the elevator at the 11th floor of a building a   [#permalink] 08 Sep 2016, 23:05
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# Steve gets on the elevator at the 11th floor of a building a

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